bzoj 1208 set，或treap或splay(待补)

1208: [HNOI2004]宠物收养所

tags：写set直接过了，但这题好像还是treap和splay的模板题，不会，留坑待补。

// set
#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a;i<=b;i++)
#define per(i,b,a) for (int i=b;i>=a;i--)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
typedef long long ll;
const int N = 2000005, mod=1000000;

int n;
ll a, b, ans;
set<ll >A, B;
int main()
{
    A.insert(-INF); A.insert(INF);
    B.insert(-INF); B.insert(INF);
    scanf("%d", &n);
    rep(i,1,n)
    {
        scanf("%lld %lld", &a, &b);
        if(a==0) {
            if(B.size()==2) A.insert(b);
            else {
                int x=*(B.upper_bound(b)), y=*(--B.upper_bound(b));
                if(abs(x-b)>=abs(y-b)) B.erase(y), ans=(ans+abs(y-b))%mod;
                else B.erase(x), ans=(ans+abs(x-b))%mod;
            }
        }
        else {
            if(A.size()==2) B.insert(b);
            else {
                int x=*(A.upper_bound(b)), y=*(--A.upper_bound(b));
                if(abs(x-b)>=abs(y-b)) A.erase(y), ans=(ans+abs(y-b))%mod;
                else A.erase(x), ans=(ans+abs(x-b))%mod;
            }
        }
    }
    printf("%lld\n", (ans+mod)%mod);

    return 0;
}